Question

A particle with charge $+Q$ and mass $m$ enters a magnetic field of magnitude $B,$ existing only to the right of the boundary $YZ.$ The direction of the motion of the particle is perpendicular to the direction of $B.$ Then, the time spent by the particle in the field will be

• A. $\frac{2\pi m}{QB}$
• B. $\frac{\pi m}{QB}$
• C. $\frac{mv}{QB}$
• D. Infinite

Answer 1

Answer: (B)

$\overrightarrow{F}=Q\left(\overrightarrow{V}\times \overrightarrow{B}\right)$
$F=Q\left(VB\right)$
Since this force is constant and is perpendicular to particle's direction of motion. Particle will move in a circle.
$\therefore QVB=\frac{m{V}^2}{R}$
$R=\frac{mV}{QB}$
$\therefore$ Time required to travel $\pi R$ distance
$t=\frac{\pi R}{V}$
$\Rightarrow t=\frac{\pi \times \frac{mV}{QB}}{v}=\frac{\pi m}{QB}$
$t=\frac{\pi m}{QB}$